A question on X validated about the probabilistic version of the mean value theorem brought me to the corresponding Wikipedia page with the explanation that, for measurable and differentiable functions g, when X and Y are positive random variables such X is stochastically dominated by Y and E[X] < E[Y], there exists a random variable Z such that
![{\displaystyle {\rm {E}}[g(Y)]-{\rm {E}}[g(X)]={\rm {E}}[g'(Z)]\,[{\rm {E}}(Y)-{\rm {E}}(X)].}](https://s0.wp.com/latex.php?latex=%7B%5Cdisplaystyle+%7B%5Crm+%7BE%7D%7D%5Bg%28Y%29%5D-%7B%5Crm+%7BE%7D%7D%5Bg%28X%29%5D%3D%7B%5Crm+%7BE%7D%7D%5Bg%27%28Z%29%5D%5C%2C%5B%7B%5Crm+%7BE%7D%7D%28Y%29-%7B%5Crm+%7BE%7D%7D%28X%29%5D.%7D&bg=000000&fg=B0B0B0&s=0&c=20201002)
What proves more interesting, however, is that this random variable has density
![f_Z(x)={\Pr(Y>x)-\Pr(X>x)\over {\rm E}[Y]-{\rm E}[X]}\,, \qquad x\geqslant 0.](https://s0.wp.com/latex.php?latex=f_Z%28x%29%3D%7B%5CPr%28Y%3Ex%29-%5CPr%28X%3Ex%29%5Cover+%7B%5Crm+E%7D%5BY%5D-%7B%5Crm+E%7D%5BX%5D%7D%5C%2C%2C+%5Cqquad+x%5Cgeqslant+0.&bg=000000&fg=B0B0B0&s=0&c=20201002)
which writes as a signed mixture of the densities proportional to Pr(Y>x) and Pr(X>x), with normalizing constants E[Y] and E[X}, and which correspond to the densities of the stationary excess variables for Y and X, respectively. (Also called residual lifetimes, which are the stationary times one need wait until the next event, when the durations between two events are distributed as Y and X, respectively.) And roughly corresponds to the extra excess time (waiting for bus Y, assuming bus X has passed in the meanwhile). A 1999 paper by Di Crescenzo provides more properties of Z, albeit not achieving a complete characterization of this random variable beyond the “stationary distribution of the difference between the current time and the last occurrence of a repair, conditional on the current state (broken or not)” in a setup with two point processes corresponding to “a repairable component that is alternately ‘up’ or ‘down’, with working and repair episodes.
![\frac{f(y)}{\int f(x)\text dx}=\frac{hf(y)}{1-\int [h(x)-h(y)f(x)/f(x)]\text dx}=\frac{hf(y)}{1-a(y)}](https://s0.wp.com/latex.php?latex=%5Cfrac%7Bf%28y%29%7D%7B%5Cint+f%28x%29%5Ctext+dx%7D%3D%5Cfrac%7Bhf%28y%29%7D%7B1-%5Cint+%5Bh%28x%29-h%28y%29f%28x%29%2Ff%28x%29%5D%5Ctext+dx%7D%3D%5Cfrac%7Bhf%28y%29%7D%7B1-a%28y%29%7D&bg=000000&fg=B0B0B0&s=0&c=20201002)
